MHB Solve Equation $\sqrt[3]{a-1}+\sqrt[3]{a}+\sqrt[3]{a+1}=0$ in Real Numbers

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The equation $\sqrt[3]{a-1}+\sqrt[3]{a}+\sqrt[3]{a+1}=0$ is solved in real numbers, leading to the conclusion that $a=0$ is the only solution. A method involving the transformation $3a = 3\sqrt[3]{a(a^2-1)}$ was discussed, resulting in $a^3 = a(a^2-1)$. Initially, there was confusion regarding the solution, but it was clarified that $a=0$ is indeed correct. Participants acknowledged calculation errors in their methods. Ultimately, the consensus is that $a=0$ is the sole solution to the equation.
anemone
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Solve in real numbers the equation $\sqrt[3]{a-1}+\sqrt[3]{a}+\sqrt[3]{a+1}=0$
 
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anemone said:
Solve in real numbers the equation $\sqrt[3]{a-1}+\sqrt[3]{a}+\sqrt[3]{a+1}=0$

using

x+y+z=0=>$x^3+y^3+x^3 = 3xyz$

we get

$(3a)^3 = 3a(a^2-1)$
a= 0 or +/-$\sqrt(1/8)$
 
kaliprasad said:
using

x+y+z=0=>$x^3+y^3+x^3 = 3xyz$

we get

$(3a)^3 = 3a(a^2-1)$
a= 0 or +/-$\sqrt(1/8)$
Neat method, but I get a different answer.

[sp]From $3a = 3\sqrt[3]{a(a^2-1)}$, I get $a^3 = a(a^2-1)$, with $a=0$ the only solution.[/sp]
 
Opalg said:
Neat method, but I get a different answer.

[sp]From $3a = 3\sqrt[3]{a(a^2-1)}$, I get $a^3 = a(a^2-1)$, with $a=0$ the only solution.[/sp]

There was a calculation mistake in my method
 
Thanks to both of you for participating and yes, $a=0$ is the only answer to the problem.:)
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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